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3 years ago

For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.

Chemistry

1 answer:

Veronika [31]3 years ago

7 0

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

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This is Q7 and I need help, can someone help.

andre [41]

Answer:

Answer to A. helium, neon, argon, krypton, xenon, and radon, B. Elemental hydrogen (H, element 1), nitrogen (N, element 7), oxygen (O, element 8), fluorine (F, element 9), and chlorine (Cl, element 17) are all gases at room temperature, and are found as diatomic molecules (H2, N2, O2, F2, Cl2). C. Elements Compounds

Ar (argon) HBr (hydrogen bromide) C 3H 8 (propane)

Kr (krypton) HI (hydrogen iodide) C 4H 10 (butane)

Xe (xenon) HCN (hydrogen cyanide)* CO (carbon monoxide)

Rn (radon) H 2S (hydrogen sulfide) CO 2 (carbon dioxide)

Explanation:

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3 years ago

In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How

____ [38]

Answer:

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution

Explanation:

First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:

Ca: 40 g/mole
F: 19 g/mole

So the molar mass of CaF₂ is:

CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?

$moles=\frac{0.0016 grams*1 mole}{78 grams}$

moles=2.05*10⁻⁵

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?

$mass of CaF_{2}=\frac{1000 mL*1g}{1mL}$

mass of CaF₂= 1000 g

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?

$moles=\frac{1000 grams*1 mole}{78 grams}$

moles=12.82

12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution

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3 years ago

What are the parts of a chromosome

juin [17]

Answer:

the centromere, the arm and the telomere

Explanation:

3 0

3 years ago

A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe

Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: $V_{1}$ = 571 mL, $T_{1} = 26^{o}C$

(a) $T_{2} = 5^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL$

(b) $T_{2} = 95^{o}F$

Convert degree Fahrenheit into degree Cesius as follows.

$(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL$

(c) $T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL$

8 0

3 years ago

Does 1 gram of phosphorus react with 6 grams of iodine to form 4 grams of phosphorus triodine in P4(s)+6I2(s)=4PI3(s)

mafiozo [28]

Answer:

Explanation:

One mole of P₄ react with six moles of I₂ and gives 4 moles of PI₃.

When one gram phosphorus and 6 gram of iodine react they gives 8.234 g ram of PI₃ .

Given data:

Mass of phosphorus = 1 g

Mass of iodine = 6 g

Mass of PI₃ = ?

Solution:

Chemical equation:

P₄ + 6I₂ → 4PI₃

Number of moles of P₄:

Number of moles = Mass /molar mass

Number of mole = 1 g / 123.9 g/mol

Number of moles = 0.01 mol

Number of moles of I₂:

Number of moles = Mass /molar mass

Number of moles = 6 g / 253.8 g/mol

Number of moles = 0.024 mol

Now we will compare the moles of PI₃ with I₂ and P₄.

I₂ : PI₃

6 : 4

0.024 : 4/6×0.024 = 0.02

P₄ : PI₃

1 : 4

0.01 : 4 × 0.01 = 0.04 mol

The number of moles of PI₃ produced by I₂ are less it will be limiting reactant.

Mass of PI₃ = moles × molar mass

Mass of PI₃ = 0.02 mol × 411.7 g/mol

Mass of PI₃ = 8.234 g

4 0

3 years ago